The de Broglie wavelength of a particle accelerated through a potential difference V is given by λ=2mqVh.
For an electron, the charge is e and mass is me. Its de Broglie wavelength is λe=2meeVh.
For a proton, the charge is e and mass is mp. Its de Broglie wavelength is λp=2mpeVh.
Taking the ratio of the two wavelengths, we get λpλe=2mpeVh2meeVh=memp.
Answer: memp